Develop Reference

File is not deleted when file path is generated dynamically

I am facing a problem but have not able to solve it yet. Let me share what I have done till now. I tried to delete a file using java.nio.file packages. And below is my code.
// directory will be dynamically generated.
String directory = fileDirectory+ "//" + fileName;
Path path = Paths.get(directory);
if (Files.exists(path)) {
Files.delete(path);
}
I generated the path correctly. But when Files.exists(path) calls it return false. That's why file is not deleted. But if I generated the directory string by hard-coded than it works perfectly.
// hard-coded directory works perfectly.
String directory = "C://opt//tomcat//webapps//resources//images//sprite.jpg";
I also tried the another method Files.deleteIfExists(path);. Which check the both the file existence and delete the file.
The other packages org.apache.commons.io.FileUtils and java.io.File have tried. But can't resolve the issue.
Note: My application is in spring-boot. And I read the directory from the application.properties file for both save images and delete images.
EDIT:
file uploading I mean save into the directory is perfectly worked. But file deletion does not work.
application.properties
image.root.dir=images
image.root.save.dir=C:/opt/tomcat/webapps/resources/
in implementation file
#Value("${image.root.dir}")
private String UPLOADED_FOLDER;
#Value("${image.root.save.dir}")
private String saveDir;
String directory = saveDir + UPLOADED_FOLDER + "/" + fileName;
save file into directory
String directory = saveDir + UPLOADED_FOLDER + "/";
try {
byte[] bytes = file.getBytes();
Path path = Paths.get(directory);
if (!Files.exists(path)) {
Files.createDirectories(path);
}
path = Paths.get(directory, file.getOriginalFilename());
Files.write(path, bytes);
} catch (IOException e) {
logger.error("save image into directory : " + e);
}

String directory = fileDirectory+ "//" + fileName;
This is not the correct separator used between a directory and a file name, though it seems to work as well.
This means the problem is not the separator, but a mismatch between the code that you use to generate the path and this code. You're generating the directory into somewhere else than where this is pointing.

Java get resource is not working

I try to read a resource file from my application but it doesn't work.
Code:
String filename = getClass().getClassLoader().getResource("test.xsd").getFile();
System.out.println(filename);
File file = new File(filename);
System.out.println(file.exists());
Output when I execute the jar-file:
file:/C:/Users/username/Repo/run/Application.jar!/test.xsd
false
It works when I run the application from IntelliJ but not when I execute the jar-file. If I open my jar-file with 7-zip test.xsd is located in the root-folder. Why isn't the code working when I execute the jar-file?

Also, File refers to actual OS file-system files; in the OS's file-system, there is only a jar file, and that jar file is not a folder. You should either extract the contents of the URL to a temporary file, or operate with its bytes in-memory or as a stream.
Note that myURL.getFile() is returning a String representation, and not an actual File. In a similar way, this will not work:
File f = new URL("http://www.example.com/docs/resource1.html").getFile();
f.exists(); // always false - will not be found in the local filesystem
A nice wrapper could be the following:
public static File openResourceAsTempFile(ClassLoader loader, String resourceName)
throws IOException {
Path tmpPath = Files.createTempFile(null, null);
try (InputStream is = loader.getResourceAsStream(resourceName)) {
Files.copy(is, tmpPath, StandardCopyOption.REPLACE_EXISTING);
return tmpPath.toFile();
} catch (Exception e) {
if (Files.exists(tmpPath)) Files.delete(tmpPath);
throw new IOException("Could not create temp file '" + tmpPath
+ "' for resource '" + resourceName + "': " + e, e);
}
}

Java path in Linux

I have a bit problem, and i dont seem to understand what is causing it.
i have a folder in my project, and in that folder i have a class, and i have a resource file (in this case jasper report).
but the only way i can access file is with absolute path or some path that starts from root of my project.
String path = "src/main/java/Views/LagerMain/lager.jrxml";
^^this works, both my class LagerController and lager.jrxml are under LagerMain folder, but when i try to do this :
String path = "lager.jrxml";
i have an error that file is not found.
I tried googling this to have a better understanding but i found nothing.
Bottom line, why cant i access my file, from class when they are both on same place, why does not relative path work.

If the main class is in a different directory, then the program will try to accesslager.jrxml there instead of the directory of the regular class.
For regular-class directory:
String path = new String(MyClass.class.getProtectionDomain().getCodeSource().getLocation()
.getPath() + System.getProperty("line.separator") + "lager.jrxml");
If that doesn't work, try this:
// your directory
File f = new File("src");
File[] matchingFiles = f.listFiles(new FilenameFilter() {
public boolean accept(File dir, String name) {
return name.startsWith("lager") && name.endsWith("jrxml");
}
});
If you have more than one file with the name lager.jrxml, then this method will return both of them and you will need to use a for to cycle through them. Otherwise, you can just use
String path = new String(matchingFiles[0].getAbsolutePath())
For main-class directory:
String path = new String(System.getProperty("user.dir")
+ System.getProperty("line.separator") + "lager.jrxml");

How to write any type of file (for example txt file) to a resources folder with the config.property file, but without using absolute path file in java

How to write or read any type of file (for example .txt file) to a resources folder with the config.property file, but without using absolute path file.
I tried to solve this like below:
ClassLoader classLoader = Setting.class.getClassLoader();
Setting setting = new Setting();
try (InputStream resourceAsStream = classLoader.getResourceAsStream("config.properties")) {
setting.load(resourceAsStream);
}
String readFileName = setting.getValue("pathSource.txt");
String writeFileName = setting.getValue("outPutPathSourceFile.txt");
String s = System.getProperty("line.separator");
File readFile = new File("./src/main/resources" + File.separator + "pathSource.txt");
File writeFile = new File("./src/main/resources" + File.separator + "outPutPathSourceFile.txt");
However, I don't want using ./src/main/resources prefix.

If your file is located under resources you able to access it like:
File file = new File(classLoader.getResource(fileName).getFile());
if you are using Spring you can use ResourceUtils.getFile():
File file = ResourceUtils.getFile("classpath:fileName")
UPDATE:
If I understand correctly, you want to read file which located under your src/main/resources without relative path usage /src/main/resources.
I created small demo:
public class ReadResourceFile {
public static void main(String[] args) throws IOException {
String fileName = "webAutomationConfig.xml";
ClassLoader classLoader = ReadResourceFile.class.getClassLoader();
File file = new File(classLoader.getResource(fileName).getFile());
System.out.println("File exists: " + file.exists());
String content = new String(Files.readAllBytes(file.toPath()));
System.out.println(content);
}
}
Output:
File exists: true
<?xml version="1.0" encoding="utf-8"?>
<config>
<baseUrl>https://www.gmail.com</baseUrl>
</config>
Project structure:

Resources (files on the class path, possibly inside a jar)can be read, but are not intended to be written to. You can use them as template to create an inital copy on the file system. – Joop Eggen

url = new java.net.URL()

url = new java.net.URL(s) doesn't work for me.
I have a string C:\apache-tomcat-6.0.29\webapps\XEPServlet\files\m1.fo and need to make a link and give it to my formatter for output, but malformed url recieved. It seems that it doesn't make my string to url.
I want also mention, that file m1.fo file is in files folder, in my webapp\product\, and I gave the full path to string like: getServletContext().getRealPath("files/m1.fo"). What I am doing wrong? How can I recieve the url link?

It is possible to get an URL from a file path with the java.io.File API :
String path = "C:\\apache-tomcat-6.0.29\\webapps\\XEPServlet\\files\\m1.fo";
File f = new File(path);
URL url = f.toURI().toURL();

Try: file:///C:/apache-tomcat-6.0.29/webapps/XEPServlet/files/m1.fo

It isn't preferable to write file:/// . Indeed it works on windows system,but in unix - there were problems.
Instead of using
myReq.put("xml", new String []{"file:" + System.getProperty("file.separator") +
getServletContext().getRealPath(DESTINATION_DIR_PATH) +
System.getProperty("file.separator") + xmlfile});
you can write
myReq.put("xml", new String [] {getUploadedFileURL (xmlfile)} );
, where
public String getUploadedFileURL(String filename) {
java.io.File filePath = new java.io.File(new
java.io.File(getServletContext().getRealPath(DESTINATION_DIR_PATH)),
filename);
return filePath.toURI().toURL().toString();

A file system path is not a URL. A URL is going to need a protocol prefix for one. To reference file system use "file:" in front of your path.